Is there a problem in this statement: As $\sum_{n=1}^{\infty} \mu(n)/n = 0, \sum_{n=1}^{N} \mu(n)/n = -\sum_{n=N+1}^{\infty} \mu(n)/n$.

Question

As by Landau's proof $$\sum_{n=1}^{\infty} \mu(n)/n = 0$$ Therefore for any $N \in \mathbb{N}$, $$ \sum_{n=1}^{N} \mu(n)/n = -\sum_{n=N+1}^{\infty} \mu(n)/n$$

Is there a problem with the above statement? A friend of mine has concerns about the conditional convergence causing issues due to Riemann's rearrangement theorem. I am of the opinion that the sequence is not being changed, just the series is being split. We aren't able to convince each other.

Thanks in advance.

Answer 1

This is always true if the series converges. It doesn't even matter if the terms are positive, the convergence is absolute, etc. Indeed, we have $$0 = \sum_{n=1}^\infty\mu(n)/n = \sum_{n=1}^N \mu(n)/n + \sum_{n=N+1}^\infty \mu(n)/n$$

Thus, the result follows by substracting $\sum_{n=N+1}^\infty \mu(n)/n$, which is allowed because the series converges by my assumption.

Answer 2

By definition, $$\sum_{k=1}^\infty a_k=L $$ means that for every $\epsilon>0$, there exists $n_0$ such that $$\left|\sum_{k=1}^n a_k-L\right|<\epsilon \text{ for all }n>n_0. $$ So if $n>\max\{n_0,n\}$, we have $$ \sum_{k=1}^na_k=\sum_{k=1}^Na_k+\sum_{k=N+1}^{n}a_k$$ and hence $$\left|\sum_{k=N+1}^{n}a_k- \left(L-\sum_{k=1} Na_k\right)\right|<\epsilon$$ for all $n>\max\{n_0,n\}$. We conclude $$ \sum_{k=N+1}^{\infty}a_k=L-\sum_{k=1}^Na_k.$$